Problem: What is the area of the region between the graphs of $f(x)=\dfrac{2x}{3\pi}-\dfrac{5}{3}$ and $g(x)=\cos(x)$ from $x=2\pi$ to $x=\dfrac{5}{2}\pi$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $6\pi-1$ (Choice B) B $\dfrac{\pi}{12}-1$ (Choice C) C $1+\dfrac{\pi}{12}$ (Choice D) D $1-6\pi$
Visualizing the area We sketch the graphs of $~f~$ and $~g~$ first. ${2}$ ${4}$ ${6}$ ${8}$ ${1}$ ${\llap{-}1}$ $f$ $g$ $y$ $x$ From the graph, it appears that $g(x)\ge f(x)$ between $x=2\pi$ and $x=\dfrac52\pi$. From this we are looking to evaluate: $ \int_{2\pi}^{5\pi/2}\left( g(x)-f(x) \right)\,dx$ Evaluating the definite integral $\begin{aligned} &\phantom{=} \int_{2\pi}^{5\pi/2} \left( \cos(x) - \left( \dfrac{2x}{3\pi}-\dfrac{5}{3} \right)\right) \,dx \\\\ &= \sin(x) -\dfrac{x^2}{3\pi}+\dfrac{5}{3}x~\Bigg|_{2\pi}^{5\pi/2} \\\\ &= \left(1- \dfrac{25\pi}{12} +\dfrac{25\pi}{6} \right) -\left( 0- \dfrac{4\pi}{3}+\dfrac{10\pi}{3} \right) \\\\ &= 1+\dfrac{\pi}{12} \end{aligned}$ Answer The area is $1+\dfrac{\pi}{12}$ square units.